In this circuit, R1 is not even required. The base current will be only Ic/β to pull the emitter within Vbe (~0.65V) below the Plus rail.

Driving the base from a port pin on something like an Arduino, the port pin will only have to source a few hundred uA.

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But it is bad design to rely on a particular value of β (or for a circuit's behavior to be heavily dependent on it). Also, the emitter will not be within Vbe of the Plus rail (by which I'm assuming you mean the +5V rail), but rather there will be the additional drop across the base resistor, which will be dependent on β.

If the base drive is a consistent voltage (doesn't have to be 5V but can't be more than about 5.5V (and that's pushing it) then you can eliminate both R1 and the base resistor and figure that, when on, the emitter will be Vbe below the base drive voltage and size the 330Ω resistor accordingly. As long as your base drive voltage is above about 3V this will work, but you do want it to be a pretty stable voltage at whatever voltage it is.

Another way to do it would be to put the transistor below the LED and 330Ω resistor and then size the base resistor so as to put the transistor into saturation when on. That will allow you to be pretty insensitive to fluctuations in the base drive voltage over a very wide range of values starting from around 1V and going up from there to even voltage well in excess of 5V. Of course, there are limits in all things.

But it is bad design to rely on a particular value of β (or for a circuit's behavior to be heavily dependent on it). Also, the emitter will not be within Vbe of the Plus rail (by which I'm assuming you mean the +5V rail), but rather there will be the additional drop across the base resistor, which will be dependent on β...

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I disagree. Look at the following sim. I model the ucontroller port as a voltage source with a series resistance. I plot the current through the series port resistor I(Rport) purple trace. Note that the peak port/base current is less than 50uA, which will not tax the port pin in the slightest. Adding an external series resistor between the port pin and the base of the emitter-follower will do nothing...

Note the LED current I(D1) red trace. It can be tailored by changing R1.

1) That it is not good to design circuits that rely on a particular transistor β.

2) That the emitter voltage will be not be the 5V minus Vbe but, rather, the 5V minus Vbe minus the voltage dropped across the base resistor.

or

3) That the voltage dropped across the base resistor will be dependent on β.

So rerun your simulation with the base resistor (which you did not say to remove, you only said that R1 was not needed) in place. A 1kΩ resistor is pretty tame and will only drop 50mV at a base current of 50uA, which represents about 2.5% of the regulating voltage (the voltage across the 150 Ω resistor). Your simulation is running with a β of about 280, near the max value. But what if the β is actually more like 80, which is not unreasonable for this collector current and Vce. Then the drop across the base resistance (the 1kΩ plus the additional 100Ω in your output model) would be about 200mV which, while far from disastrous, would not amount to a reduction in LED current of about 10%, which might well be undesirable (depends on the application and many applications would not care).

Be far more undesirable would be the variation that would be caused by a base drive voltage that was not always 5V. If this is being driven by TTL, for instance, the output level even without any load at all might is often in the 3.5V range and can be as low as 2.7V and still meet spec. Even with infinite β, that would be sufficient to go from 14mA in the LED to an LED current of only 4mA (at 3.5V, which would be a reduction in current of over 60%) to the LED not even turning on at all (at 2.7V).

I mean, wouldn't it be better if the transistor sank current rather than source it, as is the case in the examples shown in this thread? Or is this not important in this case?

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It depends on what is important and what is not and that, in turn, depends on what the circuit is being used for. If you are just trying to use the transistor as a switch to turn something one or off, then it is usually better to use a topology that allows the transistor to be taken from a cutoff condition to a saturated condition over a small range of input signal so that it behaves like a well-defined switch. This also tends to minimize the power that is dissipated in the transistor, which is sometimes a very important factor. On the other hand, if it is important that you be able to switch between on and off quickly, then you may not want to let the transistor go completely into either cutoff or into saturation as you must now bring it back out of that state, which takes time.

Sorry, I haven't used TTL to drive a LED in twenty years... and even if I was, I still wouldn't use a series resistor between the TTL output pin and the base of the emitter-follower. I would instead put a pull-up resistor between the TTL output pin and the 5V rail to source the required base current simultaneously with pulling the base within mV of the 5V rail. That gets the emitter voltage within ~700mV below the 5V rail.

I'm talking about driving a LED from a modern CMOS-totem-pole output, the kind like you find on PICs and Atmels that can source 20mA at ~4V when powered on 5V. When lightly loaded, such a CMOS output will pull within mV of the 5V rail.

If using the emitter-follower current-booster as I showed in the sim, the beta of the transistor doesn't matter, as long as (beta) X (port-pin source current) is greater than the emitter current. Even the crummiest transistor will source hundreds of mA in this configuration. The transistor is not saturated, so the usual Ib=Ic/10 rule doesn't apply.

btw- I was responding to the TS's original emitter-follower led driver; not proposing that he put the LED/resistor in the collector circuit.

Question: how come most circuits involving an NPN transistor have the load on the positive side of the circuit?

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Because with this arrangement, it's easier to raise the base voltage to 0.7V - referenced to the emitter which is at ground - to turn on the current flow. If the transistor sits on top the load in an emitter-follower arrangement, then the required base voltage is 0.7V plus the ∆V drop across the load. That ∆V might not be well known, and therefore the required base voltage is unknown. I believe a result is that the transistor is less fully turned on, and will get hotter.

I mean, wouldn't it be better if the transistor sank current rather than source it, as is the case in the examples shown in this thread? Or is this not important in this case?

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I don't think of a transistor as sourcing or sinking, merely passing, but anyway don't you have it backwards? It IS sinking the current in the common emitter configuration, and sourcing in the emitter-follower config. But whatever, the transistor only cares about the heat, not some arbitrary term describing how the current is passing.

Question: how come most circuits involving an NPN transistor have the load on the positive side of the circuit?
I mean, wouldn't it be better if the transistor sank current rather than source it, as is the case in the examples shown in this thread? Or is this not important in this case?

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Depends on what you are trying to accomplish. If driving a LED that has a Vf of say 2V from a 5V rail, then you use a current-limiting resistor that drops the other 3V. You can either use a transistor as a common-emitter switch (LED and resistor in the collector circuit) as you have shown, or as an emitter-follower as I showed.

For LED driving, I think that the emitter-follower has the advantage. One less resistor, and less current being required from the uController port pin.

The total power wasted is in either method is (5-Vf)/Iled. In the emitter-follower, some of that is wasted in the NPN. In the common-emitter, almost all of the wasted power moves to the resistor, but there is no net reduction in wasted power.

If you use a common-emitter NPN, then you need to limit the base current with a resistor, and the net base current must be much higher, putting a bigger load on the Port pin.

Sorry, I haven't used TTL to drive a LED in twenty years... and even if I was, I still wouldn't use a series resistor between the TTL output pin and the base of the emitter-follower.

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But the OP hasn't given any indication of what he is driving the base with at all, so just because you haven't used TTL doesn't mean he isn't. Like you, I also wouldn't put a series resistor in the base circuit (unless, potentially, I was concerned about protecting against certain failure modes) when using an emitter-follower configuration but your original offering didn't say anything about removing it (which I'm thinking was what you had in mind and just didn't say it). If it is stipulated that the external base resistor be removed and that we have a solid base drive supply (when sourcing a small, but variable, base current) then life is good.